A marine auxiliary direct-current power circuit supplies a constant 48\text{ V} across thr… — Armed Services Vocational Aptitude Battery (ASVAB)

A marine auxiliary direct-current power circuit supplies a constant $48\text{ V}$ across three parallel branches. The first branch has a resistance of $16\ \Omega$ , the second branch has a resistance of $24\ \Omega$ , and the third branch has a resistance of $48\ \Omega$ . What is the total current, in amperes ( $\text{A}$ ), drawn from the power supply?

Answer: 6 A

Answer

The total current drawn from the power supply is

6\text{ A}

Each parallel branch receives the full source voltage of

48\text{ V}

. Dividing the voltage by each branch resistance yields branch currents of

3\text{ A}

2\text{ A}

, and

1\text{ A}

. Summing these branch currents gives a total circuit current of

6\text{ A}

Step-by-Step Solution

Calculate the current passing through the first branch

I_1 = \frac{48\text{ V}}{16\ \Omega} = 3\text{ A}

The full source voltage of

48\text{ V}

appears across every parallel branch.

Calculate the current passing through the second branch

I_2 = \frac{48\text{ V}}{24\ \Omega} = 2\text{ A}

Applying Ohm's Law (

I = \frac{V}{R}

) with the same supply voltage across the

24\ \Omega

branch.

Calculate the current passing through the third branch

I_3 = \frac{48\text{ V}}{48\ \Omega} = 1\text{ A}

Applying Ohm's Law (

I = \frac{V}{R}

) across the

48\ \Omega

branch.

Sum all individual branch currents to determine total current

I_{\text{total}} = 3\text{ A} + 2\text{ A} + 1\text{ A} = 6\text{ A}

In a parallel circuit, the total current equals the sum of all branch currents (

I_{\text{total}} = I_1 + I_2 + I_3

Key Concept

In a parallel DC circuit, the voltage across each branch is identical to the source voltage, and the total current supplied is the sum of the individual branch currents.

Alternative Method

Find the equivalent resistance (

R_{\text{eq}}

) first using the reciprocal formula:

\frac{1}{R_{\text{eq}}} = \frac{1}{16} + \frac{1}{24} + \frac{1}{48} = \frac{3}{48} + \frac{2}{48} + \frac{1}{48} = \frac{6}{48} = \frac{1}{8}\implies R_{\text{eq}} = 8\ \Omega

. Then, calculate total current via Ohm's Law:

I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{48\text{ V}}{8\ \Omega} = 6\text{ A}

Estimated Time:1m 0s

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Answer

Step-by-Step Solution

Key Concept

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